EXAMPLE : A Predator-Prey Model . ( Course demo : Predator-Prey/ODE/2D ) − 5 u 1 ) , u ′ 1 = 3 u 1 (1 − u 1 ) − u 1 u 2 − λ (1 − e u ′ 2 = − u 2 + 3 u 1 u 2 . Here u 1 may be thought of as “fish ” and u 2 as “sharks ”, while the term − 5 u 1 ) , λ (1 − e represents “fishing”, with “fishing-quota ” λ . When λ = 0 the stationary solutions are 3 u 1 (1 − u 1 ) − u 1 u 2 = 0 ⇒ ( u 1 , u 2 ) = (0 , 0) , (1 , 0) , (1 3 , 2) . − u 2 + 3 u 1 u 2 = 0 29
The Jacobian matrix is − 5 u 1 � � 3 − 6 u 1 − u 2 − 5 λe − u 1 G u ( u 1 , u 2 ; λ ) = 3 u 2 − 1 + 3 u 1 so that � 3 � 0 G u (0 , 0 ; 0) = ; real eigenvalues 3 , -1 (unstable) 0 − 1 � − 3 � − 1 G u (1 , 0 ; 0) = ; real eigenvalues -3 , 2 (unstable) 0 2 � − 1 √ − 1 � G u (1 complex eigenvalues − 1 2 ± 1 3 3 , 2 ; 0) = ; 7 i (stable) 6 0 2 All three Jacobians at λ = 0 are nonsingular . Thus, by the IFT, all three stationary points persist for (small) λ � = 0 . 30
In this problem we can explicitly find all solutions: Family 1 : ( u 1 , u 2 ) = (0 , 0) Family 2 : λ = 3 u 1 (1 − u 1 ) u 2 = 0 , 1 − e − 5 u 1 3(1 − 2 u 1 ) = 3 ( Note that u 1 → 0 λ = lim lim 5 ) u 1 → 0 5 e − 5 u 1 Family 3 : u 1 = 1 2 3 − 1 3 u 2 − λ (1 − e − 5 / 3 ) = 0 ⇒ u 2 = 2 − 3 λ (1 − e − 5 / 3 ) 3 , These solution families intersect at two bifurcation points , one of which is ( u 1 , u 2 , λ ) = (0 , 0 , 3 / 5) . 31
0 . 75 fish 0 . 50 0 . 25 0 . 00 1 . 5 sharks 1 . 0 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 0 . 5 quota 0 . 0 Stationary solution families of the predator-prey model. Solid/dashed curves denote stable/unstable solutions. Note the bifurcations and Hopf bifurcation (red square). 32
0 . 8 0 . 6 fish 0 . 4 0 . 2 0 . 0 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 quota Stationary solution families, showing fish versus quota. Solid/dashed curves denote stable/unstable solutions. 33
Stability of Family 1 : � 3 − 5 λ � 0 G u (0 , 0 ; λ ) = ; eigenvalues 3 − 5 λ, − 1 . 0 − 1 Hence the zero solution is : unstable if λ < 3 / 5 , and stable if λ > 3 / 5 . Stability of Family 2 : This family has no stable positive solutions. 34
• Stability of Family 3 : At λ H ≈ 0 . 67 the complex eigenvalues cross the imaginary axis: − This crossing is a Hopf bifurcation , − Beyond λ H there are stable periodic solutions . − Their period T increases as λ increases. − The period becomes infinite at λ = λ ∞ ≈ 0 . 70 . − This final orbit is a heteroclinic cycle . 35
1 . 00 0 . 75 min fish, max fish 0 . 50 0 . 25 0 . 00 − 0 . 25 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1 . 0 quota Stationary (blue) and periodic (red) solution families of the predator-prey model. ( For the periodic solution family both the maximum and minimum are shown. ) 36
Periodic solutions of the predator-prey model. The largest orbits are close to a heteroclinic cycle. 37
The bifurcation diagram shows the solution behavior for (slowly) increasing λ : • Family 3 is followed until λ H ≈ 0 . 67 . • Periodic solutions of increasing period until λ = λ ∞ ≈ 0 . 70 . • Collapse to trivial solution (Family 1). 38
Continuation of Solutions Parameter Continuation Suppose we have a solution ( u 0 , λ 0 ) of G ( u , λ ) = 0 , as well as the derivative u 0 . ˙ Here u ≡ d u ˙ dλ . We want to compute the solution u 1 at λ 1 ≡ λ 0 + ∆ λ . 39
"u" u 1 du u (= d λ at λ 0 ) � � ���� ���� u � ���� ���� 0 (0) u 1 ���� ���� � ∆λ ����������������� ����������������� λ λ λ 0 1 Graphical interpretation of parameter-continuation. 40
To solve the equation G ( u 1 , λ 1 ) = 0 , for u 1 (with λ = λ 1 fixed) we use Newton’s method G u ( u ( ν ) 1 , λ 1 ) ∆ u ( ν ) − G ( u ( ν ) = 1 , λ 1 ) , 1 ν = 0 , 1 , 2 , · · · . u ( ν +1) = u ( ν ) + ∆ u ( ν ) . 1 1 1 As initial approximation use u (0) = u 0 + ∆ λ ˙ u 0 . 1 If G u ( u 1 , λ 1 ) is nonsingular , and ∆ λ sufficiently small then this iteration will converge [B55]. 41
After convergence, the new derivative u 1 is computed by solving ˙ G u ( u 1 , λ 1 ) ˙ u 1 = − G λ ( u 1 , λ 1 ) . This equation is obtained by differentiating G ( u ( λ ) , λ ) = 0 , with respect to λ at λ = λ 1 . Repeat the procedure to find u 2 , u 3 , · · · . NOTE : • ˙ u 1 can be computed without another LU -factorization of G u ( u 1 , λ 1 ) . • Thus the extra work to compute u 1 ˙ is negligible . 42
EXAMPLE : The Gelfand-Bratu Problem . u ′′ ( x ) + λ e u ( x ) = 0 for x ∈ [0 , 1] , u (0) = 0 , u (1) = 0 . We know that if λ = 0 then u ( x ) ≡ 0 is an isolated solution . Discretize by introducing a mesh , 0 = x 0 < x 1 < · · · < x N = 1 , x j − x j − 1 = h , (1 ≤ j ≤ N ) , h = 1 /N . The discrete equations are : u j +1 − 2 u j + u j − 1 + λ e u j = 0 , j = 1 , · · · , N − 1 , h 2 with u 0 = u N = 0 . 43
Let u 1 u 2 u ≡ . · u N − 1 Then we can write the discrete equations as G ( u , λ ) = 0 , where G : R N − 1 × R → R N − 1 . 44
Parameter-continuation : Suppose we have λ 0 , u 0 , and u 0 . ˙ Set λ 1 = λ 0 + ∆ λ . Newton’s method : G u ( u ( ν ) 1 , λ 1 ) ∆ u ( ν ) − G ( u ( ν ) = 1 , λ 1 ) , 1 u ( ν +1) = u ( ν ) + ∆ u ( ν ) , 1 1 1 for ν = 0 , 1 , 2 , · · · , with u (0) = u 0 + ∆ λ ˙ u 0 . 1 After convergence compute u 1 ˙ from G u ( u 1 , λ 1 ) ˙ u 1 = − G λ ( u 1 , λ 1 ) . Repeat the procedure to find u 2 , u 3 , · · · . 45
Here − 2 h 2 + λe u 1 1 h 2 1 − 2 1 h 2 + λe u 2 h 2 h 2 G u ( u , λ ) = . . . . . . . 1 − 2 h 2 + λe u N − 1 h 2 Thus we must solve a tridiagonal system for each Newton iteration. NOTE : • The solution family has a fold where parameter-continuation fails ! • A better continuation method is “pseudo-arclength continuation ”. • There are also better discretizations, namely collocation , as used in AUTO . 46
Pseudo-Arclength Continuation This method allows continuation of a solution family past a fold . It was introduced by H. B. Keller (1925-2008) in 1977. Suppose we have a solution ( u 0 , λ 0 ) of G ( u , λ ) = 0 , u 0 , ˙ as well as the normalized direction vector ( ˙ λ 0 ) of the solution family. Pseudo-arclength continuation consists of solving these equations for ( u 1 , λ 1 ) : G ( u 1 , λ 1 ) = 0 , u 0 � + ( λ 1 − λ 0 ) ˙ � u 1 − u 0 , ˙ λ 0 − ∆ s = 0 . 47
"u" u 1 � � �� �� u 0 �� �� � � � , λ 0 ) ( �� �� u 0 u 0 � � ∆ s � � λ 0 � λ λ λ 0 1 Graphical interpretation of pseudo-arclength continuation. 48
Solve the equations G ( u 1 , λ 1 ) = 0 , u 0 � + ( λ 1 − λ 0 ) ˙ � u 1 − u 0 , ˙ λ 0 − ∆ s = 0 . for ( u 1 , λ 1 ) by Newton’s method : G ( u ( ν ) 1 , λ ( ν ) ( G 1 u ) ( ν ) ( G 1 λ ) ( ν ) 1 ) ∆ u ( ν ) � � . 1 = − ∆ λ ( ν ) ˙ � u ( ν ) u 0 � + ( λ ( ν ) − λ 0 )˙ u ∗ ˙ λ 0 1 − u 0 , ˙ λ 0 − ∆ s 0 1 1 Compute the next direction vector by solving � ˙ G 1 G 1 0 u λ � u 1 , = ˙ λ 1 ˙ u ∗ ˙ λ 0 1 0 and normalize it. 49
NOTE : u 1 , ˙ • We can compute ( ˙ λ 1 ) with only one extra backsubstitution . • The orientation of the family is preserved if ∆ s is sufficiently small. + ˙ u 1 � 2 λ 2 • Rescale the direction vector so that indeed � ˙ 1 = 1 . 50
FACT : The Jacobian is nonsingular at a regular solution. � � u ∈ R n +1 . PROOF : Let x ≡ λ Then pseudo-arclength continuation can be simply written as G ( x 1 ) = 0 , � x 1 − x 0 , ˙ x 0 � − ∆ s = 0 , ( � ˙ x 0 � = 1 ) . x 1 � �� �� x 0 x 0 ∆ s � Pseudo-arclength continuation. 51
The pseudo-arclength equations are G ( x 1 ) = 0 , � x 1 − x 0 , ˙ x 0 � − ∆ s = 0 , ( � ˙ x 0 � = 1 ) . The Jacobian matrix in Newton’s method at ∆ s = 0 is � G 0 � x . x ∗ ˙ 0 N ( G 0 At a regular solution x ) = Span { ˙ x 0 } . � G 0 � x We must show that is nonsingular at a regular solution. x ∗ ˙ 0 52
G 0 � � x If on the contrary is singular then for some vector z � = 0 we have : x ∗ ˙ 0 G 0 x z = 0 , � ˙ x 0 , z � = 0 , Since by assumption N ( G 0 x ) = Span { ˙ x 0 } , we have z = c ˙ x 0 , for some constant c . But then x 0 � 2 � ˙ x 0 , z � c � ˙ x 0 , ˙ x 0 � c � ˙ 0 = = = = c , so that z = 0 , which is a contradiction . QED ! 53
EXAMPLE : The Gelfand-Bratu Problem . Use pseudo-arclength continuation for the discretized Gelfand-Bratu problem. Then the matrix � � � � G x G u G λ = , ˙ x ∗ u ∗ ˙ ˙ λ in Newton’s method is a bordered tridiagonal matrix : ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ . ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ which can be decomposed very efficiently . 54
Following Folds and Hopf Bifurcations • At a fold the the behavior of a system can change drastically. • How does the fold location change when a second parameter varies ? • Thus we want the compute a locus of folds in 2 parameters. • We also want to compute loci of Hopf bifurcations in 2 parameters. 55
Following Folds Treat both parameters λ and µ as unknowns , and compute a solution family X ( s ) ≡ ( u ( s ) , φ ( s ) , λ ( s ) , µ ( s ) ) , to G ( u , λ, µ ) = 0 , F ( X ) ≡ G u ( u, λ, µ ) φ = 0 , � φ , φ 0 � − 1 = 0 , and the added continuation equation u 0 � + � φ − φ 0 , ˙ φ 0 � + ( λ − λ 0 )˙ � u − u 0 , ˙ λ 0 + ( µ − µ 0 ) ˙ µ 0 − ∆ s = 0 . As before, u 0 , ˙ φ 0 , ˙ ( ˙ λ 0 , ˙ µ 0 ) , is the direction of the family at the current solution point ( u 0 , φ 0 , λ 0 , µ 0 ) . 56
EXAMPLE : The A → B → C Reaction . ( Course demo : Chemical-Reactions/ABC-Reaction/Folds-SS ) The equations are − u 1 + D (1 − u 1 ) e u 3 , u ′ = 1 − u 2 + D (1 − u 1 ) e u 3 − Dσu 2 e u 3 , u ′ = 2 − u 3 − βu 3 + DB (1 − u 1 ) e u 3 + DBασu 2 e u 3 , u ′ = 3 where 1 − u 1 is the concentration of A , u 2 is the concentration of B , u 3 is the temperature , α = 1 , σ = 0 . 04 , B = 8 , D is the Damkohler number , β is the heat transfer coefficient . 57
A stationary solution family for β = 1 . 20. Note the two folds and the Hopf bifurcation . 58
1 . 4 1 . 2 1 . 40 1 . 0 1 . 35 0 . 8 β β 0 . 6 1 . 30 0 . 4 1 . 25 0 . 2 0 . 0 1 . 20 0 . 00 0 . 05 0 . 10 0 . 15 0 . 20 0 . 14 0 . 15 0 . 16 0 . 17 0 . 18 0 . 19 0 . 20 D D A locus of folds (with blow-up) for the A → B → C reaction. Notice the two cusp singularities along the 2-parameter locus. ( There is a swallowtail singularity in nearby 3-parameter space. ) 59
7 6 5 || u || 4 3 2 1 0 . 12 0 . 14 0 . 16 0 . 18 0 . 20 0 . 22 D Stationary solution families for β = 1 . 20 , 1 . 21 , · · · , 1 . 42. ( Open diamonds mark folds, solid red squares mark Hopf points. ) 60
Following Hopf Bifurcations The extended system is f ( u , λ, µ ) = 0 , F ( u , φ , β, λ ; µ ) ≡ f u ( u , λ, µ ) φ − i β φ = 0 , � φ , φ 0 � − 1 = 0 , where R n × C n × R 2 × R R n × C n × C , F : → and to which we want to compute a solution family ( u , φ , β , λ , µ ) , with u ∈ R n , φ ∈ C n , β, λ, µ ∈ R . Above φ 0 belongs to a “reference solution ” ( u 0 , φ 0 , β 0 , λ 0 , µ 0 ) , which normally is the latest computed solution along a family. 61
EXAMPLE : The A → B → C Reaction . ( Course demo : Chemical-Reactions/ABC-Reaction/Hopf ) The stationary family with Hopf bifurcation for β = 1 . 20 . 62
2 . 0 1 . 5 β 1 . 0 0 . 5 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 D The locus of Hopf bifurcations for the A → B → C reaction. 63
7 6 5 4 || u || 3 2 1 0 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 D Stationary solution families for β = 1 . 20 , 1 . 20 , 1 . 25 , 1 . 30 , · · · , 2 . 30, with Hopf bifurcations (the red squares) . 64
Boundary Value Problems Consider the first order system of ordinary differential equations u ′ ( t ) − f ( u ( t ) , µ , λ ) = 0 , t ∈ [0 , 1] , where u ( · ) , f ( · ) ∈ R n , µ ∈ R n µ , λ ∈ R , subject to boundary conditions b ( · ) ∈ R n b , b ( u (0) , u (1) , µ , λ ) = 0 , and integral constraints � 1 q ( · ) ∈ R n q . q ( u ( s ) , µ , λ ) ds = 0 , 0 65
This boundary value problem (BVP) is of the form F ( X ) = 0 , where X = ( u , µ , λ ) , to which we add the continuation equation � X − X 0 , ˙ X 0 � − ∆ s = 0 , where X 0 represents the latest solution computed along the family. In detail , the continuation equation is � 1 � u ( t ) − u 0 ( t ) , ˙ u 0 ( t ) � dt � µ − µ 0 , ˙ µ 0 � + 0 ( λ − λ 0 )˙ + λ 0 − ∆ s = 0 . 66
NOTE : • In the context of continuation we solve this BVP for ( u ( · ) , λ , µ ) . • In order for problem to be formally well-posed we must have n µ = n b + n q − n ≥ 0 . • A simple case is n q = 0 , n b = n , for which n µ = 0 . 67
Discretization: Orthogonal Collocation Introduce a mesh { 0 = t 0 < t 1 < · · · < t N = 1 } , where h j ≡ t j − t j − 1 , (1 ≤ j ≤ N ) , Define the space of (vector) piecewise polynomials P m as h � [ t j − 1 ,t j ] ∈ P m } , P m ≡ { p h ∈ C [0 , 1] : p h � h � where P m is the space of (vector) polynomials of degree ≤ m . 68
The collocation method consists of finding µ ∈ R n µ , p h ∈ P m h , such that the following collocation equations are satisfied : p ′ h ( z j,i ) = f ( p h ( z j,i ) , µ, λ ) , j = 1 , · · · , N , i = 1 , · · · , m , and such that p h satisfies the boundary and integral conditions . The collocation points z j,i in each subinterval [ t j − 1 , t j ] , are the (scaled) roots of the m th-degree orthogonal polynomial (Gauss points 3 ). 3 See Pages 261 , 287 of the Background Notes on Elementary Numerical Methods. 69
0 1 t t t t 0 1 2 N t t j-1 j � � � z j,1 z j,2 z j,3 t j-1 t j t t j-2/3 j-1/3 (t) (t) l l j,3 j,1 The mesh { 0 = t 0 < t 1 < · · · < t N = 1 } , with collocation points and extended-mesh points shown for m = 3 . Also shown are two of the four local Lagrange basis polynomials . 70
Since each local polynomial is determined by ( m + 1) n , coefficients, the total number of unknowns (considering λ as fixed) is ( m + 1) n N + n µ . This is matched by the total number of equations : collocation : m n N , ( N − 1) n , continuity : constraints : n b + n q ( = n + n µ ) . 71
Assume that the solution u ( t ) of the BVP is sufficiently smooth. Then the order of accuracy of the orthogonal collocation method is m , i.e. , = O ( h m ) . � p h − u � ∞ At the main meshpoints t j we have superconvergence : = O ( h 2 m ) . max j | p h ( t j ) − u ( t j ) | The scalar variables λ and µ are also superconvergent . 72
Implementation For each subinterval [ t j − 1 , t j ] , introduce the Lagrange basis polynomials { ℓ j,i ( t ) } , j = 1 , · · · , N , i = 0 , 1 , · · · , m , defined by m t − t j − k � ℓ j,i ( t ) = m , m − t j − k t j − i k =0 ,k � = i m where i t j − i m ≡ t j − m h j . The local polynomials can then be written m � p j ( t ) = ℓ j,i ( t ) u j − i m . i =0 With the above choice of basis u j ∼ u ( t j ) and u j − i m ∼ u ( t j − i m ) , where u ( t ) is the solution of the continuous problem. 73
The collocation equations are ′ p j ( z j,i ) = f ( p j ( z j,i ) , µ , λ ) , i = 1 , · · · , m, j = 1 , · · · , N . The boundary conditions are b i ( u 0 , u N , µ , λ ) = 0 , i = 1 , · · · , n b . The integral constraints can be discretized as N m � � ω j,i q k ( u j − i m , µ , λ ) = 0 , k = 1 , · · · , n q , j =1 i =0 where the ω j,i are the Lagrange quadrature weights . 74
The continuation equation is � 1 µ 0 � + ( λ − λ 0 ) ˙ � u ( t ) − u 0 ( t ) , ˙ u 0 ( t ) � dt + � µ − µ 0 , ˙ λ 0 − ∆ s = 0 , 0 where ( u 0 , µ 0 , λ 0 ) , is the previous solution along the solution family, and µ 0 , ˙ ( ˙ u 0 , ˙ λ 0 ) , is the normalized direction of the family at the previous solution . The discretized continuation equation is of the form N m � � ω j,i � u j − i m − ( u 0 ) j − i ( ˙ m � m , u 0 ) j − i j =1 i =0 µ 0 � + ( λ − λ 0 ) ˙ + � µ − µ 0 , ˙ λ 0 − ∆ s = 0 . 75
Numerical Linear Algebra The complete discretization consists of m n N + n b + n q + 1 , nonlinear equations , with unknowns m } ∈ R mnN + n , µ ∈ R n µ , { u j − i λ ∈ R . These equations are solved by a Newton-Chord iteration . 76
We illustrate the numerical linear algebra for the case n = 2 ODEs , N = 4 mesh intervals , m = 3 collocation points , n b = 2 boundary conditions , n q = 1 integral constraint , and the continuation equation. • The operations are also done on the right hand side , which is not shown. • Entries marked “ ◦ ” have been eliminated by Gauss elimination. • Entries marked “ · ” denote fill-in due to pivoting . • Most of the operations can be done in parallel . 77
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • The structure of the Jacobian . 78
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • • The system after condensation of parameters, which can be done in parallel . 79
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ • • • • The preceding matrix, showing the decoupled • subsystem . 80
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • Stage 1 of the nested dissection to solve the decoupled • subsystem. 81
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • Stage 2 of the nested dissection to solve the decoupled • subsystem. 82
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • • • • • • • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • The preceding matrix showing the final decoupled • subsystem . 83
u 1 µ λ u 0 u 2 u 1 u 2 u 3 u N 3 3 • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • • • ◦ ◦ ◦ ◦ • • · · • • • • ◦ ◦ ◦ ◦ ◦ • · · • • • • • • • • • • • • • • ◦ • • • • • • • • • ◦ ◦ • • • • • • • • ◦ ◦ ◦ • • • • • A A B B ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • A A ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ B B • • • • • • • • • • • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • • • • The approximate Floquet multipliers are the eigenvalues of M ≡ − B − 1 A . 84
Accuracy Test The Table shows the location of the fold in the Gelfand-Bratu problem, for 4 Gauss collocation points per mesh interval, and N mesh intervals . N Fold location 2 3.5137897550 4 3.5138308601 8 3.5138307211 16 3.5138307191 32 3.5138307191 85
Periodic Solutions • Periodic solutions can be computed efficiently using a BVP approach. • This method also determines the period very accurately. • Moreover, the technique can compute unstable periodic orbits. 86
Consider u ( · ) , f ( · ) ∈ R n , u ′ ( t ) = f ( u ( t ) , λ ) , λ ∈ R . Fix the interval of periodicity by the transformation t t → T . Then the equation becomes u ( · ) , f ( · ) ∈ R n , u ′ ( t ) = T f ( u ( t ) , λ ) , T , λ ∈ R . and we seek solutions of period 1 , i.e. , u (0) = u (1) . Note that the period T is one of the unknowns . 87
The above equations do not uniquely specify u and T : Assume that we have computed ( u k − 1 ( · ) , T k − 1 , λ k − 1 ) , and we want to compute the next solution ( u k ( · ) , T k , λ k ) . Then u k ( t ) can be translated freely in time : If u k ( t ) is a periodic solution, then so is u k ( t + σ ) , for any σ . Thus, a phase condition is needed. 88
An example is the Poincar´ e phase condition ′ � u k (0) − u k − 1 (0) , u k − 1 (0) � = 0 . ( But we will derive a numerically more suitable integral phase condition . ) u k-1 (0) u k-1 (0) � � u (0) k Graphical interpretation of the Poincar´ e phase condition. 89
An Integral Phase Condition If ˜ u k ( t ) is a solution then so is u k ( t + σ ) , ˜ for any σ . We want the solution that minimizes � 1 u k ( t + σ ) − u k − 1 ( t ) � 2 D ( σ ) ≡ � ˜ 2 dt . 0 The optimal solution u k ( t + ˆ ˜ σ ) , must satisfy the necessary condition D ′ (ˆ σ ) = 0 . 90
Differentiation gives the necessary condition � 1 u ′ � ˜ u k ( t + ˆ σ ) − u k − 1 ( t ) , ˜ k ( t + ˆ σ � dt = 0 . 0 Writing u k ( t ) ≡ ˜ u k ( t + ˆ σ ) , gives � 1 � u k ( t ) − u k − 1 ( t ) , u ′ k ( t ) � dt = 0 . 0 Integration by parts, using periodicity, gives � 1 ′ 0 � u k ( t ) , u k − 1 ( t ) � dt = 0 . This is the integral phase condition. 91
Continuation of Periodic Solutions • Pseudo-arclength continuation is used to follow periodic solutions . • It allows computation past folds along a family of periodic solutions. • It also allows calculation of a “vertical family ” of periodic solutions. For periodic solutions the continuation equation is � 1 u k − 1 ( t ) � dt + ( T k − T k − 1 ) ˙ T k − 1 + ( λ k − λ k − 1 )˙ � u k ( t ) − u k − 1 ( t ) , ˙ λ k − 1 = ∆ s . 0 92
SUMMARY : We have the following equations for periodic solutions : u ′ k ( t ) = T f ( u k ( t ) , λ k ) , u k (0) = u k (1) , � 1 ′ � u k ( t ) , u k − 1 ( t ) � dt = 0 , 0 with continuation equation � 1 u k − 1 ( t ) � dt + ( T k − T k − 1 ) ˙ T k − 1 + ( λ k − λ k − 1 )˙ � u k ( t ) − u k − 1 ( t ) , ˙ λ k − 1 = ∆ s , 0 where u ( · ) , f ( · ) ∈ R n , λ , T ∈ R . 93
Stability of Periodic Solutions In our continuation context, a periodic solution of period T satisfies u ′ ( t ) = T f ( u ( t )) , for t ∈ [0 , 1] , u (0) = u (1) , (for given value of the continuation parameter λ ). A small perturbation in the initial condition u (0) + ǫ v (0) , ǫ small , leads to the linearized equation v ′ ( t ) = T f u ( u ( t ) ) v ( t ) , for t ∈ [0 , 1] , which induces a linear map v (0) → v (1) , represented by v (1) = M v (0) . 94
v (1) = M v (0) The eigenvalues of M are the Floquet multipliers that determine stability. M always has a multiplier µ = 1 , since differentiating u ′ ( t ) = T f ( u ( t )) , gives v ′ ( t ) = T f u ( u ( t ) ) v ( t ) , where v ( t ) = u ′ ( t ) , with v (0) = v (1) . 95
v (1) = M v (0) • If M has a Floquet multiplier µ with | µ | > 1 then u ( t ) is unstable . • If all multipliers (other than µ = 1) satisfy | µ | < 1 then u ( t ) is stable . • At folds and branch points there are two multipliers µ = 1 . • At a period-doubling bifurcation there is a real multiplier µ = − 1 . • At a torus bifurcation there is a complex pair on the unit circle. 96
EXAMPLE : The Lorenz Equations . ( Course demo : Lorenz ) These equations were introduced in 1963 by Edward Lorenz (1917-2008) as a simple model of atmospheric convection : x ′ = σ ( y − x ) , y ′ = ρ x − y − x z , z ′ = x y − β z , where (often) σ = 10 , β = 8 / 3 , ρ = 28 . 97
Course demo : Lorenz/Basic Bifurcation diagram of the Lorenz equations for σ = 10 and β = 8 / 3 . 98
Course demo : Lorenz/Basic Unstable periodic orbits of the Lorenz equations. 99
Recommend
More recommend
